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VARIATION BY AMPLITUDE.

1. Turn the given time into astronomical time; if P. M., by putting down the given day, hour, and minute; if A. M., by adding 12 to the hours, and placing before them a day less than the date at ship.

2. Find the apparent time at Greenwich, by adding the long. if W., and subtracting if E.

3. With the Greenwich time, get the declination from page 1 of the month, and correct it.

4. Put down the latitude and the cor. dec. under it; take out the log. secant of the lat. and the log. sine of the dec.; add these two logs. together; the sum is the log. sine of the true amplitude.

5. The true amplitude is to be named always from the E. if A. M., and from the W. if P. M., and towards the N. or S. as the cor. declination.

6. Put the magnetic or observed amplitude under the true ; add them together if different names, but subtract them if the

same names.

7. The result is the variation; E. if the true is to the right, and W. if to the left, of the magnetic.

EXAMPLE I. 1870, June 17th, at 8h. 12m. P. M. apparent time at ship, in latitude 51° 28' N., longitude 40° W., the magnetic amplitude of the Sun was W. 27° 45' N. Required the true amplitude, and the variation of the compass.

d. h.
Long. 40° W. App. time at ship 17 12
4
Long. in time

2 40 W.

m.

8

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46.870

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EXAMPLE II. 1870, May 12th, at 5h. 32m. A. M., apparent time at ship, in latitude 20° 10' N., longitude 30° W., the magnetic amplitude of the Sun was E. S. Required the true amplitude, and the variation of the compass.

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LONGITUDE BY CHRONOMETER. 1. To the time by chronometer apply the error given; adding it if slow, but subtracting it if fast. Then apply the accumulated rate; adding if losing, but subtracting if gaining. This will give the true Greenwich mean time.

Note.—This accumulated rate is found by multiplying

the number of days elapsed from the date of the error being given, to the date of the question, by

the daily rate. 2. With the Green. time, take out the declination and equation of time from the Nautical Almanac, page 2 of the month, and correct both in the usual way. Then get the polar distance by adding the declination to 90° 00'00”, if the latitude and declination are different names; but subtracting it from 90°, if they are the same names. 3. Correct the observed alt. for I. E. and for Table IX.

Add together the true altitude, latitude, and polar distance, divide their sum by 2, and subtract the true altitude from this sum.

5. Take out the secant of the latitude, cosecant of the polar distance, cosine of the } sum, and sine of the remainder, and add the four logs. together; their sum (rejecting all tens) is the log. of the hour angle found in Table XXXI. The following form shews this work.

True alt.
Lat.

Secant 0.
Polar dist.

Cosecant 0.
2) Sum

Cosine
Remainder

Sine
Hour angle

Table XXXI. 6. Put down the apparent noon at ship, which is the day given 00h. 00m. 00s.; under it put the hour angle; and if Ă. M. at ship, always subtract the hour angle from apparent noon; if P. M., add. This is the apparent time at ship. Apply the equation of time, as directed at the top of page 1 of the Nautical Almanac. The result is the mean time at ship.

7. Put the mean time at Greenwich under the mean time at ship, and the difference between them is the longitude in time, which turned into degrees and minutes, is the longitude of ship; East, if Gr. time is the less ; West, if it is the greater.

Note.-If the polar distance is greater than 90°, the cosecant of it is the same as the secant of the corrected declination,

D

Add.

# Sum

EXAMPLE I. 1870, March 10th, P. M. at ship, in latitude 40° 10' S., the observed alt. of the Sun's l. 1. was 27° 10' 30", I. E. + 1' 15", height of the eye 14 feet; when a chronometer shewed 10d. 4h. 17m. 20s., which was slow for Green. mean time 3m. 195. on January 21st, and losing 23.5 daily. Find the longitude.

d. h. m.

From Jan. 21 Time by chron.

} = 48 days. 10 4 17 20

2.5 daily rate. Error, slow

3 19 10 4 20 39

240

96 Loss in interval + 2 00 True Gr. mean time 10 4 22 39

6,0) 12,0.0

s.

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S.

111 days.

EXAMPLE II. 1870, August 27th, P. M. at ship, in latitude 15° 30' S., the observed alt. of the Sun's 1. 1. was 20° 49' 50", I. E. — 2' 10", height of the eye 16 feet; when a chronometer shewed 26d. 19h. 57m. 24s., which was fast for Green. mean time lh. 10m. 37s., on May 7th, and gaining 35-7 daily. Find the longitude.

d. h. m. From May 7 Time by chron. 26 19 57 24 to Aug. 26

3.7 daily rate. Error, fast

1 10 37 26 18 46 47

333 Gain in interval

6 53 True m. time at Gr. 26 18 39 54

12 | | 410.7
6 1.85

.92

6,0) 41,3-47 Gain in interval 6 53

hours.

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16 12 Obs. alt. 20° 49' 50% True alt. 20° 57' 10" I. E.

2 10

Lat. 15 30 00 Secant 0.016089 20 47 40

P. dist. 100 08 22 Cosecant 0.006828 Cor. + 9 30

2) 136 35 32 True alt, 20 57 10

Sum

68 17 46 Cosine 9:567904 Remainder 47 20 36 Sine 9.866586 Hour angle 4 18 59 (XXXI.) 9.457407

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d. h. m. Apparent noon at ship 27 00 00 00 Hour angle

4 18 59 (P. M. add.) Apparent time at ship 27 4 18 59 Equation of time

+

27 Mean time at ship

27 4 20 26 Mean time at Greenwich 26 18 39 54 Longitude in time

9 40 32

60

4 ) 580 32 Longitude 145° 8' East.

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